structural analysis

7/16/2019 Structural Analysis

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Advanced Structural Analysis

Prof. Devdas Menon

Department of Civil Engineering

Indian Institute of Technology, Madras

Module No. # 5.3

Lecture No. # 29

Matrix Analysis of Beams and Grids

Good morning. This is lecture number 29, module 5, Matrix analysis of beams and grids.

Today, we will complete the application of conventional stiffness method and we will

introduce the reduced elements stiffness method.

(Refer Slide Time: 00:32)

This is covered in this book, which you need to refer to Advanced Structural Analysis.

We will continue with conventional stiffness method, remember we did different types of

problems. We first took a non-prismatic fixed beam. Then, we did continuous beam,


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remember three span continuous beam, which also had an overhang and we did not take

any shortcuts. We have dealt with different boundary conditions; we also dealt with

internal hinges, we also dealt with intermediate supports. Now, here is some more types

of supports that you can get you can get the extreme end, as hinged or roller and you can

get a guided fixed support.

Now, it is possible to take advantage of reduced degree of kinematic indeterminacy. We

will do that in the reduced elements stiffness method, but in the conventional stiffness

method, we do not take any shortcuts. Actually, it is quiet easy to deal with any situation,

you have to assign the appropriate boundary conditions.


In the case of a hinged end support, as you see at the end A in this example, we have an

active rotational degree of freedom, along with a restrained translational deflection

degree of freedom. At support C, guided fixed end support, we have an active

translational degree of freedom, along with a restrained rotational degree of freedom.


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We have done this example earlier, remember, it is a symmetric continuous beam; we

can take advantage of symmetry; we cut the beam at the middle at C and and insert, what

kind of support, do we insert there and guided fixed, guided roller support, as shown

here. And let us see, how to analyze this.


What are the active degrees of freedom? How many of them do we have? Yeah. three.

three.


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three active and the method of analysis exactly the same, so 1 2 3 we start with the left

end rotation at A, rotation at B, deflection at C- these are the three active degrees of

freedom.

Then, the restrain degrees of freedom are 4 5 6, as indicated there. As far as the local

coordinates are concerned, just you have two beam elements, each has four degrees of

freedom and if you had to write down the In this particular problem, do we have any

any nodal loads? No, they are all distributed loads, there is no support settlement given

now.

You have to write down the transformation matrix. What does it look like? Can you try

the two transformation matrices? Well, they are identity matrices. You just have to

assign the linking global coordinates.

What are the linking global coordinates for number one?

5 1

6 2

You have to do it in that order: 5 1 6 2, for element one: for element two:

6 2 3 4

6 2 3 4. You must know the sequence, because the the translation must match, with the

translation. Do not mix up the rotation.


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So, is this, ok 5 1 6 2, 6 2 3 4 - that is all-, very easy to do.


Next, you have to find the fixed end forces. We have done this problem earlier, so I

think we can go quiet fast. You can work out the fixed end forces. Mind you we are not

taking shortcuts, so you are arresting all degrees of freedom that you identified, as active

degrees of freedom.

For example, in the second case, C is arrested. The the rotation at C is arrested and these

are simple fixed end moment calculations, that you can do. If you recall, we did this


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problem by slope deflection and moment distribution method. There, we took a shortcut;

we took advantage of the fact, that, there was a guided roller support there and we

modified both the stiffness and the fixed end forces.

We do not take those shortcuts in conventional stiffness method. We will take that

shortcut, in reduced element stiffness method, because this is, the normal way

computers, software solve these problems. We just want to stimulate, what the standard

computer program does, is it clear.

So, we really do not care, how many degrees of freedom there- are, we just want to

correctly identify the degree of kinematic indeterminacy. We do not want to take

shortcuts, of the kind that we did earlier. But, if there is symmetry in the structure,

definitely it is worth taking, dealing with half the frame, so you can work out these.

These are very straight forward.

(Refer Slide time: 05:59)

Under constrained load, the formula is minus, not minus, because now it is plus minus-

since- you are dealing with the conventional stiffness method, where anticlockwise is

positive. So, it is W a b squared by L squared, you know that formula, you can apply all

that and you can get the fixed end force vectors.

One more important difference here, you need to get the vertical reactions as well, which

is something- we did not do, when we did slope deflection and moment distribution


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method. Those are easy to compute, in the first case, since, it is symmetric. Just take the

total load, 15 into 4 divided by 2, you get the two reactions. When you have an un-

symmetric load, you get the shears; not only from the load, but you also get from the lack

of balance, between these two loads.

In this case, because this should this should be both should be anticlockwise. This should

be anticlockwise and the difference, divided by the span, you should add up.

Yeah, this should be clockwise, you are right. This should be clockwise, it is an error

shown there but the vectors are correctly shown. Pre-multiply, why do we do this

exercise? Because, we want to find out, what the equivalent joint forces are in the

structure.


So, you are pre-multiplying by an identity matrix but you get along with it the linkingcoordinates and then you just you have done the slotting; you just assemble them

properly. You have six degrees of freedom: three are active, three are restrained.

Corresponding to one- you have 20 kilonewton meter, corresponding to two- you have

minus 20 and plus 22.5, which add up, 2.5 five and so on and so forth. So, this is this can

be easily assembled. Is it clear?

This is your fixed end force vector. What do we do next? You find the net, the resultant

load vector, which is FA minus Ff. FA in this case, is zero, it is a null vector, because you


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do not have any nodal loads, in this particular problem and so, you can find the resultant

force. You can even, draw a sketch to tell, exactly what is happening.


This is your resultant force vector and to remind you once again, the displacements you

get, at the active degrees of freedom: D1, D2 and D3 from this problem. From this, nodal

load problem will be exactly equal to, the displacements you would have got, with theoriginal intermediate loads. Is it clear? That is the equivalence, we are trying to establish.

It is very effectively done, because you can do matrix analysis, only if you have dealing

with nodal loads. Because, matrix had only vectors, which have discrete entities, so your

coordinates are limited.


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Next, you generate the element and structure stiffness matrices. Element stiffness matrix

is standard, there, you have two elements. All you need is EI and L. In the first case,

actually EI is constant, for both the elements. L is 4 meters for first element, 2 meters for

the second element. So, plug in those values, you can get kstar1and kstar2. Very simple,

straight forward and we have done this problem earlier by moment distribution method.



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What do we do, next? We have to assemble, the Structure stiffness matrix. To do that,

we have to convert, from the local axis system, to the global axis system, through the

transformation- Ti transpose kstar into Ti. .

So, you do this first and then you, actually, you get the same thing again, when you pre-

multiply by Ti transpose, you got the linking coordinates, then, you have to do the

slotting, which I will not explain. You will finally, end up with a, full 6 by 6 matrix: the

top upper corner is always kAA, the bottom lower corner is kRR the upper corner is kAR

and the lower is kRA.

You should write a program ideally which will generate this automatically. You should

just take a look and check there, it is a symmetric and it is the diagonal elements are all

positive. You are You are set to handle any loading. this And this is what the computer

does, the moment you give, the geometry of the structure and the material properties, it is

ready, with the stiffness matrix and then it is waiting for the loads. You also got the

loads.

Now, you these are your equilibrium equations. If you solve the first equation, you see

how familiar this whole thing is now, nothing. I mean once, you have got the method;

you got the solution.



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To solve the first one and you will get some solution, in this problem, EI was not

mentioned. So, if you plug in the value of EI, you will get the answer in millimeters and

radians.

What do you do next? Find the support reactions, how do you do that? Take the second

equilibrium equation and plug in the value of DA in this problem, DR is the null vector.

You will get some answers and what should you do after getting these answers? No,

before that, you must check equilibrium; you must check equilibrium and it is a simple

check; you can do many checks; you can do moment equilibrium check. The least you

can do is, you just add up the total downward forces, must add up to total upward forces

and it is exact to the to the third decimal place, in your solution.


Then you find the member forces, you are familiar with this, here you have to be careful

about one thing, what is that? You got there is a F stari f -missing. What is that you need

to be careful about?

The Di you have to be careful, because if you go back, you got displacement, D1 D2D3.

When you go to the element coordinates, you have four degrees of freedom here. You

must get the correct Direlevant to to the element that you are taking into consideration.

For example, the first element, this is a fixed end force vector, which you have already

got, this is kstar1T1 which you already calculated and stored in your computer. Now, you


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recognize, there are linking coordinates of 5 1 6 2. So, you have to put D5 D1 D6 and D2,

some of them are restrained.

For example, 5 and 6 are restrained, so they are zero. You have to pick up D 1 and D2,

from the solution that you generated from your DA vector. This is the only thing, you

have to be careful about, at the rest of it follows. Is it clear?

You have to select those displacements, from your final displacement vector, which has

DA and DR, which are matching which are which are matching you are an element and

that comes nicely, with the linking, coordinates which come along with kstari Ti.


Do a similar exercise, for the second element and you get some moments and then you

can draw your free body diagram, bending moment diagram. This is exactly the same

solution, we got by the methods, we did earlier. Is it clear?


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One last topic, before we move onto, reduced element stiffness method. Sometimes, it

becomes necessary, to account for, shear deformations. when When do you think, you

need to count for, shear deformations?

The length is very small.

When, the span to depth ratio is is low.

In normal beams, the span to depth ratio is, how much? Well, 10 to 15. If it is above 8 or

7, you would not make any significant error, by ignoring shear deformations and

normally, well proportioned beam belong to that category. Occasionally, for various

reasons, possibly for architectural reasons, you might require, require a deep beam. So, if

you got a deep beam, even if you do not have a deep beam; if you still include shear

deformations, youre only going to get a more accurate solution.

So, that deep beams are those beams, in which the span to depth ratio is roughly less than

two and half, between two and half and six or seven, you have the the the middle range,

which is between, shallow beams and deep beams.

You encounter such situations, especially, when you deal with shear walls, in tall

buildings. So, you know shear walls are the lateral load resisting system, a part of the

lateral load resisting system, in a multi storey building. The shear wall is essentially, avertical cantilever but, it is connected to different floors, through the horizontal framing


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plan. You you may have beams; you may have slabs and essentially, if, it subject to and

the lateral loads come from, those diaphragm is at the slab levels, in the different storied

building of a structure.

Now, if the shear wall is slender, in the sense, it is it is like this, (Refer Slide Time:

17:24) width is small and the stories height is large, there are many stories and

essentially, it is going to behave, like a flexural element.

It will be the deflection shape, it will be like that. But, sometimes you encounter shear

walls, which are very wide and stocky, the height is less. So, when you try to push that

shear wall; youll find that, the deflection that you get, has two components: one is

flexural, the other is shear. Remember, we did this exercise earlier.

Shear deflection in cantilever beams and the shear deflection can be comparable to the

flexural deflection. In a cantilever subject to a constant load p at the free end, what is the

deflection? p L cube by 3 EI. But, there is also a shear deflection that you get.

Remember, we did this, which has g a dash coming into the picture, in the denominate.

So, in such instances, shear deformations have to come in. Now, this kind of beam,

which you include shear deformations was was originally, described in detail, by

Timoshenko and sometimes it referred to as Timoshenko beam.

It is an energy formulation. I am not giving the formulation here, but you can go through

it in literature. You will find that, you have to modify, your element stiffness matrix, by

including a quantity, where you have a ratio of the flexural rigidity, to the shear rigidity.

Now, that is called, the shear deformation constant betas.


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So, you have EI and you have G A dash, I, A dash. G is the shear modulus, which is

related to the elastic modulus, through the Poissons ratio. You know that.

A dash, is not the gross cross sectional area, it is the reduced cross sectional area, to

account for the fact that, the shear stress distribution, is not uniform. .

So, there is a form factor that comes into play. So, this is a non-dimensional factor, that

is why, L square also comes in the denominator and we will see, how whats the

variation of this factor? How it affects the analyses? But, you are and sometimes this

formula is useful. You can easily run this and check, what is the error I make, if, I ignore

shear deformations; by including shear deformations, in any beam problem.

This can be also used, when you do frame problems. So, all you need to do, is to modify

this term outside, by this factor: 1 plus betas in the denominator and wherever you have 4

EI by L, you have to add 4 plus betas, into this constant; where you had 2, you have to

put minus betas, that is all you need to do.

So, four terms, in your coefficient matrix get affected and all terms get affected, because

this is common outside. Is it, clear? So, this is a correction, you need to do and check out

the solution.


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Considering, a rectangular section and a material with Poissons ratio, mu equal to 0.3,

taking on values in the practical range of 0.01, for very high span to depth ratios, 2 3.12,

for L by D equal to 1. Normally, you would not get a depth of beam, which is less than

the span of the beam. So, the variation of 1 divided by 1 plus beta s is between 1 and 0.24

in practice.

So, that is that is a kind of range, that you get obviously when 1 by 1 plus betas is goes to

0.24, it is going to make a big difference but, those are very rare on exceptional cases.

We are not going to do any problem but, if tomorrow you are need to include it, in a real

life situation, please make use of this matrices. With this, we have covered completely,

the conventional stiffness method.


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Now, we will look at the reduced element stiffness method. In this reduced. In the

conventional stiffness method, we have this stiffness matrix. I am not including shear

deformations.

What is the rank of this matrix?

3

3

2

3

2. Rank is 2. It is not 3; it is 2. You can easily make out. Is it, not? Take a look. You can

play around with rows and columns and two of them, will you know find the reduced

form but more easily, you can see, why is it? Why is the rank not 4 but, 2? What, is it,

0.2?

There are only two equations of equilibrium are available, that is why, you cant you

cannot find the inverse of this matrix. It is a singular matrix, but in physically, what does

it, mean? when you When you find, the rank is not full? What is it, point two?

It depends equations, sir. The other forces physically


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What does it? We have discussed this earlier.

The other forces are depend..

It is unstable. It can have rigid body motion. See, you can have a flexibility matrix, onlyfor a stable structure. You can have stiffness matrix for an unstable structure that, at the

element level, not structure. At the structures to be stable, otherwise your structure

stiffness matrix will become singular. This is very interesting. The element stiffness

matrix is singular but, your structural stiffness matrix will not be singular, unless you

take the full k matrix. You have kAA. kAA will not be singular. Because, you are arresting

some degrees of freedom and you have to arrest a certain minimum number, for it to be

stable. Is it clear?

Now, what is the to make this stable, what should, we do? You can arrest, any of these

two degrees of freedom. If you make it simply supported, it is going to be stable. So, that

is the element: we will use; we did this for axial element, remember. We caught hold of

one end, we said, we can pull only the other end.


Here also, you arrested, when you have a simply supported beam. So, from now on, in

the reduced element stiffness method; this is the the this is the element, that we will use

for the element.


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You can, it is a good question. Can you use a cantilever? Please, do so, most of us like,

simply supported conditions and we are already familiar with, you know 2 star and 4

star. The 4 EI by L 2, EI by L, is now, childs play, for all of us. So, for shear

convenience, we have used simply support, but you can also use, cantilever. In the book,

I have discussed this. In the book, I have also used cantilever as an option, but I do not

want to confuse you. So, go by simply supported, but you can also use this.

Now, can you write down, the element stiffness matrix, for this, straight away? It is a 2

by 2 matrix and we will use a symbol k tilda, what is it?


Can you extracted from the the full, four by four matrix? Can how How do you, extract

it? What should you remove? what should you remove? What should you remove, from

that, 4 by 4 matrixes?

The translation column.

Translation column.

Just knock off the first column, third column, first row, third row and whatever you left

with. That is an easiest, you can do. Whatever, you are left with, is your reduced element

stiffness. Let us do it in, a way, that you will appreciate better, let us, use a physical

approach and derive the same thing.


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So, what do you normally do, to get the first column? You apply D 1 star equal to 1 and

the moment link with that is 4 EI by L, you have a carryover moment at the other end,

which is 2 EI by L. That is all, you need to worry about. Do not worry, about the

reactions.

If you want to, the deflected shape will look like that; the point of contra flexural willseparate the beam and the ratio two isto one. Bending moment diagram will look like

that.

If you now, do D 2 star equal to 1, you get a similar figure and from this, you can

generate, your element stiffness matrix. Is it, clear? Very easy, to re and very easy to

remember: 4 EI by L; 2 EI by L; 2 EI by L; 4 EI by L, no minus signs, here. Clear? easy.


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Now, you can also do it from flexibility matrix, because you get, if you invert the

flexibility matrix, you get the reduced element stiffness matrix.

In an axial element, it was very easy, because it was 1 by 1. It was EA by L or L by A.

Here, it is not 1, it is 2 by 2. Let us, do this, you remember. If I take at the same simply

supported beam, apply fstar1 equal to 1, the deflected shape will look like this, if I useconjugate beam method or some such method, I can show that this angle here, fstar1 is L

by 3 EI and the angle here will be half this, put it in the opposite direction, it is minus L

by 6 EI. If I do, F 2 star equal to 1, it look like this. So, what is my flexibility matrix? It

is going to, look like this. Here, is negative sign comes.

If I take the inverse of this matrix, I get that matrix. You can work both ways; you can

generate the flexibility matrix, from the stiffness matrix or you can, but, you can do it

only, with the reduced element stiffness. Is it clear?

Now, let me ask you a simple question. How can you use a two degree of freedom

element? How can it handle the deflections, in your structure?

Remember, your global coordinate system: It has not only rotation; it also has

deflections: with this 2 by 2, can you handle it? It is an important question, to ask? Is it

clear?


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How do we, How did we manage in slope deflection and moment distribution method?

Chord rotation. That is right, we use chord rotation. Normally, if you do not have chord

rotation, your interpretation in this element AB is, MAB is Fstar1, MBA is Fstar2, theta A is

Dstar1 and theta B is Dstar2: these are flexural rotations.

If you have chord rotation, that means you have relative movement, differentialsettlement. Now, we switch, our sign conventions, we say that anticlockwise chord

rotation is positive; Vertical reflection upward is positive, not downward. So, DB minus

DA both is acting upward. If DB is more than DA, we will have an anti clockwise

rotation and phi will be positive. Is it clear?

How do you deal, with this? What happens in a beam, when you have an anticlockwise

chord rotation? You get clockwise end moments, so how how can we take advantage, of

this? Your stiffness matrix does not change. So, how do you accommodate, that is right.

you have to you cannot touch the left hand side. You play with the Dstar. You have to

somehow accommodate, the chord rotation in your in your, displacement vector, how do

you do that? So, you have to do it in the, displacement transformation matrix. So, how do

you do that?

Theta a minus 3i by.

Theta A minus 5i. So, that is what, you do.


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So, it is a clever trick. If you have an anticlockwise chord rotation, just, recognize that

the effect it has, is that of a clockwise flexural rotation at the both the ends.


Let us demonstrate, this once, because most people, who have done this, have really not

fully understood this. I hope, you understand, why we are doing it? You can go back to

our earlier sessions and see how we did it.

So, let us say, you need, you have some, displacement transformation matrix to generate

for for this element. Let us do it, component by component; let us say, from you global


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coordinate system, you have some Dj, that Dj could be a translation; could be a rotation;

usually, there will upward or anti clockwise. Got it? You want to write down, the effect

of that, on this two degree of freedom element.

There are four possibilities. Let us look at them. Two possibilities are shown here, there

are pretty easy. Let us say in your, continues beam structure, you had a rotation Dj

anticlockwise. It will either, go to the left or it will go to the right.

So, do you, agree. If Dj corresponds to the left end rotation, obviously D1 star will be one

time Dj and D2 star will be zero. If Dj corresponds to the, right side of your element,

anticlockwise, then it will be zero one. This is clear. This is straightforward. There is

nothing special about it, the problem comes, when you have a deflection. So, let us look

at that. Let us say you had a deflection, in your structure, in your, continues beam that

means your element is going to deflect like this, all other coordinates, all other

displacements being arrested. So, how will you write, D1 star D 2 star for this, if length of

the element is end.

See, I wrote something here, you have to write something similar here. For this situation,

what will you write here? What is the first component? What is the second component?

Dj by L

NO

0 minus Dj by L 1 minus Dj by L

No

No, Dj is a deflection.

1 by L is your chord rotation. 1 by L is a, because Dj is a deflection. You do not put

chord rotation in the global coordinates; you put only deflections. Get it straight.

So, you put 1 by L, now, the question, should you put plus 1 by L or minus 1 by L? So,

you have to think, if the left end is going up, then you having a clockwise chord rotation,

so you have anticlockwise end moments. So, you should put, plus, because anticlockwise

is positive. Well done. The last case is, when the right end goes up, that is it. Is it clear?


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Just to seal, your understanding, let us take a look at this situation. I have shown an

element here, which is a part of continues beam. The coordinates are l m n p, it could be

anything. So, this part of continues beam and this could be active degrees of freedom,

restrained, I do not care. So, they let us say, they come together like this-D l Dm Dn Dp, at

the global level, at the element level I have only 1 star and 2 star. The element

coordinates are D1 star, D 2 starand you need to do this transformation. Can you write down

T Di, in terms of L?

So, we are talking of this segment, you will have four terms: 1 2 3 4. Tell me, what

those four terms? Are you getting it? What are those four terms?

1 by L 1 0 2by L

one erratum

1 by L 1 by L 1

1 0 1 by L

this is this portion This portion, let us try to figure out.

It is very easy. If I lift this up, that is my first coordinates, D l by, unity. Do, I get a

clockwise chord rotations?


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Yes, sir.

Do, i get a clockwise chord rotations?

Yes, yes, sir.

So, I get anticlockwise end moments, which are positive.

So, the equivalent rotation is positive. Got it?

Number two: If I put Dm starequal to 1, what do I get? I get a unit rotation, on the left end.

So, its remains positive and I do not get anything here, D 2 staris zero.

Then I have Dn stargoing up by unity, what is my chord rotation? Anticlockwise. So, what

are my end moments? Clockwise. Clockwise means, it is minus.

So, do you get the hang of it? That is it. Because, once you have got this, we can proceed

ahead. There you can deal with any problem, but it is only, this little bit, you need to

understand, the procedure is the same, you have to generate, these matrices. The

procedure is identical.


Let us demonstrate quickly, with one example. This example, which we have already

done, by the conventional stiffness method.


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So, find out the T D matrix, manually generates the fixed end force vector. Remember,

we had problems, when you had distributed loads in the axial system. Why, because it

can handle only a constant one force. Here, you have a distributed loading, so you, you

the two end moments, may may not be the same.


We will demonstrate this and the other rest of the procedure same, fixed end forces isonly, manually supreme force or generally.So, coordinate transformation, if you look

at global coordinates, these are the same coordinates; we used in conventional stiffness

method; we are not changing here. Same coordinates and local coordinates are what we

have just discussed two degrees of freedom, write down the transformation matrix.


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Write down the transformation, there are two matrices; you have to write TD1 and TD2.

You can to put altogether, also.

So, let me help you, first of all the given loading is, you have a nodal moment 30

kilonewton meter. So, this is what you get. Let us do it together. Listen carefully. D 1

equal to 1, this goes up, it causes an anticlockwise chord rotation in element one andcauses a clockwise chord rotation in element two, so when, you have anticlockwise

chord rotation, you will get, clockwise end moment. So, you should put a minus sign.

So, it is minus 1 divided by 10, so you get minus 1 divided by 10, for the left one and for

the right one, you get plus 1 divided by 10. Is this clear? You all got it.

Am just extending what we did, so you fill up the first column, for which covers both T D

1 and TD 2, second one D 2 equal to 1, you have a rotation here. The right end of this willhave 1, the left end of this will have 1.

So, 1 1 0, 1 1 0. Is it clear? Clear.

Now, you move to the restrain coordinates, 3 equal to 1, if you lift this up, it affects only

element one. You have a clockwise chord rotation element one, so anticlockwise that is

positive. So, you get plus 0.1, plus 0.1, nothing happens to the second element.


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Number four: If you give a rotation A it effects only D1 star1 in the first element, nothing

else happen to 1 0 0 0.

Number five: You lift this up, you have anticlockwise chord rotation in the second

element, clockwise end moments, so equivalent flexure rotations will be negative 0.1,

0.1, in the second element, nothing happens. Is this clear? and so on, so far.

Can you do this, correctly? You have done. So it is a powerful technique of dealing with

deflections, converting them to equivalent flexural rotations, through the displacement

transformation matrix. We do not touch the stiffness matrix, so if you got this step,

correct? It is easy.


Fixed end forces, we have done earlier. So, you can find the fixed end forces, find the

deflections, if you want, we have done this, earlier. Put the fixed end force, (Refer SlideTime: 42:30) in a vector form. You got this diagram, so, this you have to directly get. In

conventional stiffness method, you could assemble it, doing the contra gradient principle.


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You can put, Ti transpose. Here, you cannot do that. Here, you have to do it manually,

that means you draw these sketches and then you figure out. I want to find out F 1. So, F1

will be 50 plus 50, 100. Is it clear? I want to find out F 2. F2, is this coordinate that will

be, 83.33 minus 125. So, minus 46.

I want to find F3, that is, this force that is plus 50. I want to find F4, that is plus 125, anticlockwise I want to find F5 , that is plus 50. I wanted to find F6 that is (( )). Is it clear?

You have to do it by inspection. No, contra gradient principle, when you are using the

reduced elements stiffness method, but this is ok, in fact this method is, more for manual

use.

Yeah, that is nice. We also, do it here by the way. Here also, we write F star. the this

does not change but, we have only 2 here. There we had a 4 by 1, we including 50 50also. We did the Ti transpose and we got the linking coordinates and then we did slotting.

It was all done mechanically, here you cannot do that. You have done it with your eyes,

open. Is it clear?


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Then find the net load vector, from now on, its same as conventional stiffness method.

Because, you you got beyond the element, so this part is common and of course the

element stiffness matrix derivation, you have to do carefully.

So, you have two elements, span is same, El is changing. Very easy to write down: 4 2 4

2, 4 2 2 4 EI by L. What did you do after you get a K star, you do K star TD. TD you havealready got, do not have to write any linking coordinates, because of all the global

coordinates, in one go. Then, what did you do? You pre- multiply this with TD. You just

sum up, you got the same structure stiffness matrix, which you generated, in the

conventional stiffness method.

You manage with less coordinates, then this solution is, ditto, same as what you did in

conventional stiffness method. Because, you have to invert the K, you have to find so,

that we are just reproducing, what we did earlier.

What you do next? Find the support reactions; same equation. You can check

equilibrium. Then last step, member forces you have to do little carefully, you have that

you have done this, you remember, when you computed the structure stiffness matrix,

you first did K i star TD i, that comes in handy here. Because, that is what you inverse

here. .


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And plug in those values will get the same answers. Please, go through this problem

solved in the book also, exactly same solution. Is it clear? So, we demonstrated one

application, we will do all the other applications, as well in the next class. Thank you.

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structural analysis

Documents

method of analysis

support c

fixed end support

guided fixed support

beam elements

matrix analysis of beams

guided roller support

hinged end support